Evaluate $\int\frac{dx}{(3+4\sin x)^2}$

Question

Evaluate: $$\int\frac{dx}{(3+4\sin x)^2}$$

My attempt:

I have tried to express the integrand in terms of $\tan x$ and $\sec x$ but there was no use since the substitution $\tan x=z$ is of no use after that. I also tried to use the Weierstrass substitution, but I got a very complicated algebraic expression.

Please help.

Answer 1

Hint: I think the general solution for these types integrals is Tangent half-angle substitution, with $$\sin x=\dfrac{2t}{1+t^2}~~~,~~~dx=\dfrac{2}{1+t^2}\ dt$$ the integral simplifies to $$\int\frac{dx}{(3+4\sin x)^2}=\int\frac{2}{(3t^2+8t+3)^2}\ dt$$ then the squaring of denominator gives the result.

Answer 2

Hint:

Integrating by parts,

$$\int\dfrac{\cos x\ dx}{\cos x(a+b\sin x)^n}=\dfrac1{\cos x}\int\dfrac{\cos x\ dx}{(a+b\sin x)^n}-\int\left(\dfrac{d(\sec x)}{dx}\int\dfrac{\cos x\ dx}{(a+b\sin x)^n}\right)dx$$

$$=\dfrac{}{b(1-n)\cos x(a+b\sin x)^{n-1}}-\int\dfrac{\sin x}{b(1-n)(1-\sin^2x)(a+b\sin x)^{n-1}}$$

Here $n=2$

Now use Partial fraction, $$\dfrac{\sin x}{(1-\sin^2x)(a+b\sin x)}=\dfrac A{1+\sin x}+\dfrac B{1-\sin x}+\dfrac C{a+b\sin x}$$

and Weierstrass substitution in the last integral as the first two are elementary.

Answer 3

Integrate both sides of the equation $$\left( \frac{4\cos x}{3+4\sin x}\right)’ = -\frac{3}{3+4\sin x}-\frac{7}{(3+4\sin x)^2} $$to reduce the integral $$I=\int\frac{dx}{(3+4\sin x)^2} = -\frac{4\cos x}{7(3+4\sin x)}- \frac{3}{7} \int \frac{dx}{3+4\sin x} $$ and note that $\int \frac{dx}{3+4\sin x} =\int\frac{d(\frac{4+3\sin x}{\cos x})}{(\frac{4+3\sin x}{\cos x})^2-7} $. Thus $$I= -\frac{4\cos x}{7(3+4\sin x)}+ \frac{3}{7\sqrt7} \coth^{-1}\frac{4+3\sin x}{\sqrt7\cos x} $$

Answer 4

By integration by parts, we have $$ \begin{aligned} \int \frac{d x}{(3+4 \sin x)^{2}} =&-\frac{1}{4} \int \frac{1}{\cos x} d\left(\frac{1}{3+4 \sin x}\right) \\ =&-\frac{1}{4 \cos x(3+4 \sin x)}+\frac{1}{4} \underbrace{\int \frac{\sec x \tan x}{3+4 \sin x}}_{J} d x \end{aligned} $$ $$ J=\int \frac{\sec x \tan x}{3+4 \sin x} d x= \int \frac{\sin x}{(1+\sin x)(1-\sin x)(3+4 \sin x)} d x $$

Resolving the integrand of the last integral yields $$ J=\frac{1}{2} \underbrace{ \int \frac{d x}{1+\sin x} }_{K} +\frac{1}{14}\underbrace{\int \frac{d x}{1-\sin x}}_{L}-\frac{12}{7} \underbrace{\int \frac{d x}{4 \sin x+3}}_{M} $$ $$ \begin{aligned} K&=\int \frac{1-\sin x}{\cos ^{2} x} d x=\int\left(\sec ^{2} x-\tan x \sec x\right) d x=\tan x-\sec x+C_{1} \end{aligned} $$ Similarly, $$ L=\int \frac{1+\sin x}{\cos ^{2} x} d x=\tan x+\sec x+C_{2} $$

Letting $t=\tan \frac{x}{2}$, $$ \begin{aligned} M &=\int \frac{1}{4\left(\frac{2 t}{1+t^{2}}\right)+3} \cdot \frac{2 d t}{1+t^{2}}=2 \int \frac{d t}{3 t^{2}+8 t+3} =\frac{2 \sqrt{7}}{7} \ln \left|\frac{3 t-\sqrt{7}+4}{3 t+\sqrt{7}+4}\right|+C_{3} \end{aligned} $$

Now we can conclude that $$ \begin{aligned} I=&-\frac{1}{4 \cos x(3+4 \sin x)}+\frac{\tan x}{7}-\frac{3 \sec x}{28} +\frac{6 \sqrt{7}}{49} \ln \left|\frac{3 \tan \frac{x}{2}+\sqrt{7}+4}{3 \tan \frac{x}{2}-\sqrt{7}+4}\right|+C\\=& \frac{6 \sqrt{7}}{49} \ln \left|\frac{3 \tan \frac{x}{2}+\sqrt{7}+4}{3 \tan \frac{x}{2}-\sqrt{7}+4}\right|-\frac{4 \cos x}{7(3+4 \sin x)}+C \end{aligned} $$